Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Kelley-Namioka "Linear Topological Spaces",Page 31, it gives two equivalent definitions of projective topology(i.e.,weak topology). I don't know why they are equivalent.

The point is I don't kown why the following statement is equivalent to the definition of the weak topology:

"The projective topology(A.K.A. the weak opology) can be described alternatively by specifying that a subset $U$ of $X$ is open relative to te projective topolgy if and only if $U$ is the inverse under $P$ of an open subset of $\prod_{f\in F}{Y_f}$, where $P$ is the map which sends a point $x$ of $X$ into the point with $f$-th coordinate $f(x)$(that is, $P(x)(f)=f(x)$)"

I hope someone can tell me how to prove it.

share|improve this question
    
To clarify the question to those who don't have access to Kelley-Namioka: This is about the initial topology of a family $F$ of maps $f \colon X \to Y_f$ from a set $X$ to topological spaces $Y_f$. The question is why the initial topology (the weakest topology making all $f$ continuous) can be equivalently described as the weakest topology on $X$ making the evaluation map $P \colon X \to \prod_{f\in F} Y_f$ continuous. –  t.b. Sep 15 '12 at 2:10
    
No,$P$ is not a evaluation map, it maybe a representation of x. $P$ should be in $X^{**}$. –  Chao Sep 15 '12 at 2:18
    
We are only talking about topological spaces, there are no dual spaces here, so $X^{\ast\ast}$ makes no sense. The map $P \colon X \to \prod_{f \in F} Y_f$ is called the evaluation map on Wikipedia because it evaluates the functions $f$ simultaneously. $P \colon x \mapsto (f(x))_{f \in F} \in \prod_{f \in F} Y_f$. –  t.b. Sep 15 '12 at 2:21
    
If you think my comments are incorrect and that I misunderstood your question then please add both definitions to your question in order to make this question self-contained. –  t.b. Sep 15 '12 at 2:31
2  
Very good :) Welcome to the site! –  t.b. Sep 15 '12 at 8:38

1 Answer 1

up vote 2 down vote accepted

Suppose that $F$ is a family of maps from a set $X$ to spaces $Y_f$ indexed by $F$. Let $Y=\prod_{f\in F}Y_f$. Let $$P:X\to Y:x\mapsto\langle f(x):f\in F\rangle$$ be the evaluation map. One way to show that the weakest topology making each $f\in F$ continuous is the same as the weakest topology making $P$ continuous is to show that a topology $\tau$ on $X$ makes $P$ continuous iff it makes each $f\in F$ continuous.

Suppose that $\tau$ is a topology on $X$ making each of the maps $f$ continuous, so that for each $f\in F$ and each open set $U\subseteq Y_f$, $f^{-1}[U]\in\tau$. Then $U$ is the union of basic open sets of the form

$$B=\{y\in Y:\forall f\in F_0(y_f\in U_f)\}\;,\tag{1}$$

where $F_0$ is a finite subset of $F$, and $U_f$ is an open set in $Y_f$ for each $f\in F_0$. Thus, to show that $P^{-1}[U]\in\tau$, it suffices to show that $P^{-1}[B]\in\tau$ for each $B$ of the form $(1)$. But

$$\begin{align*}x\in P^{-1}[B]&\text{ iff }P(x)\in B\\ &\text{ iff }P(x)_f\in U_f\text{ for each }f\in F_0\\ &\text{ iff }f(x)\in U_f\text{ for each }f\in F_0\\ &\text{ iff }x\in\bigcap_{f\in F_0}f^{-1}[U_f]\;, \end{align*}$$

so $$P^{-1}[B]=\bigcap_{f\in F_0}f^{-1}[U_f]\;.$$

By hypothesis each of the sets $f^{-1}[U_f]\in\tau$, so $P^{-1}[B]\in\tau$ as well, and so is any union of such sets.

Conversely, suppose that $\tau$ makes $P$ continuous. Then $P^{-1}[B]\in\tau$ for every basic open set in $Y$ of the form $(1)$. In particular, for each $f\in F$ and any open set $U_f\subseteq Y_f$, $B=\{y\in Y:y_f\in U_i\}$ is open in $Y$, and therefore $P^{-1}[B]\in\tau$. But

$$\begin{align*} P^{-1}[B]&=\{x\in X:P(x)\in B\}\\ &=\{x\in X:P(x)_f\in U_f\}\\ &=\{x\in X:f(x)\in U_f\}\\ &=f^{-1}[U_f]\;, \end{align*}$$

so $f^{-1}[U_f]=P^{-1}[B]\in\tau$.

This shows that the topologies making $P$ continuous are exactly the same as the topologies making every map in $F$ continuous, and it follows immediately that the weakest topology making $P$ continuous is the weakest topology making every map in $F$ continuous.

share|improve this answer
    
Thanks for your proof, i've got it. –  Chao Sep 15 '12 at 6:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.